Question 977060
According to the time-honored algebra conventions you were taught as "order of operations",
sqrt(x) + sqrt(y) / sqrt(x) - sqrt(y) ={{{sqrt(x)+sqrt(y)/sqrt(x)-sqrt(y)}}}
I bet what you meant is
( sqrt(x) + sqrt(y) ) / ( sqrt(x) - sqrt(y) ) ={{{(sqrt(x)+sqrt(y))/(sqrt(x)-sqrt(y))}}} ,
with parentheses to indicate that the numerator and denominator need to be calculated first, before dividing.
To rationalize, you multiply both parts (numerator and denominator) times the "conjugate" of the denominator.
(The conjugate is almost the same thing, but with the sign changed).
It works because {{{(a-b)(a+b)=a^2-b^2)}}} ,
so with {{{a=sqrt(x)}}} and {{{b=sqrt(y)}}} , you have
{{{(sqrt(x)-sqrt(y))(sqrt(x)+sqrt(y))=(sqrt(x))^2-(sqrt(y))^2=x-y}}} ,
and your denominator ends up with no square roots.
{{{(sqrt(x)+sqrt(y))/(sqrt(x)-sqrt(y))}}}={{{((sqrt(x)+sqrt(y))/(sqrt(x)-sqrt(y)))((sqrt(x)+sqrt(y))/(sqrt(x)+sqrt(y)))}}}={{{(sqrt(x)+sqrt(y))^2/((sqrt(x)-sqrt(y))(sqrt(x)+sqrt(y)))}}}={{{(sqrt(x)+sqrt(y))^2/((sqrt(x))^2-(sqrt(y))^2)}}}={{{( (sqrt(x))^2 + (sqrt(y))^2 + 2sqrt(x)sqrt(y))/(x-y)}}}={{{(x+2sqrt(xy)+y)/(x-y)}}}=( x + 2sqrt(xy) + y ) / (x-y)
That is what you tried to write for choice a.
What you wrote is
 x + 2sqrt(xy) + y / x - y ={{{x + 2sqrt(xy) +y / x - y }}}